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A thin uniform rod has mass 'M' and length 'L'. The moment of inertia about an axis perpendicular to it and passing through the point at a distance $\frac{\mathrm{L}}{3}$ from one of its ends, will be
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$\frac{\mathrm{ML}^{2}}{9}$
(B)
$\begin{aligned} \mathrm{I} &=\mathrm{I}_{0}+\mathrm{Mh}^{2} \\ \mathrm{~h} &=\frac{\mathrm{L}}{2}-\frac{\mathrm{L}}{3}=\frac{\mathrm{L}}{6} \\ \mathrm{I} &=\frac{\mathrm{ML}^{2}}{12}+\mathrm{Mh}^{2} \\ &=\frac{\mathrm{ML}^{2}}{12}+\frac{\mathrm{ML}^{2}}{36} \\ &=\frac{4 \mathrm{ML}^{2}}{36}=\frac{\mathrm{ML}^{2}}{9} \end{aligned}$
$\begin{aligned} \mathrm{I} &=\mathrm{I}_{0}+\mathrm{Mh}^{2} \\ \mathrm{~h} &=\frac{\mathrm{L}}{2}-\frac{\mathrm{L}}{3}=\frac{\mathrm{L}}{6} \\ \mathrm{I} &=\frac{\mathrm{ML}^{2}}{12}+\mathrm{Mh}^{2} \\ &=\frac{\mathrm{ML}^{2}}{12}+\frac{\mathrm{ML}^{2}}{36} \\ &=\frac{4 \mathrm{ML}^{2}}{36}=\frac{\mathrm{ML}^{2}}{9} \end{aligned}$
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